michel123456 Posted December 19, 2010 Share Posted December 19, 2010 (edited) Let's take a rod. A hollow cylinder. Let's put it horizontaly, with a laser device at one end. The rod is standing still at rest. Let's open the laser for a time T so that light travels inside the cylinder a distance D. In Spacetime, light goes from point A to B, having travelled the interval X. Something like fig.1 here below: In this diagram, speed of light, C, is the angle of the line AB. It is expressed as a fraction: meters/seconds. Because we are measuring light, we can pose that [math] D=CT [/math] (1) From the diagram, using Pythagoream theorem, we can pose that [math] X^2 = D^2 + T^2 [/math] (2) Replacing (1) in (2) we get [math] X^2 = C^2 T^2 + T^2 [/math] (3) From the diagram, it appears that X corresponds to the definition of the Minkowski spacetime interval S. Because it is a lightlike spacetime interval, S=0* If we pose that X (from the diagram) is the same as S (from Theory) we get [math] X^2 = C^2 T^2 + T^2=S=0[/math] (4) or [math]C^2 T^2 + T^2=0[/math] (5) which has the only solution [math]C^2=-1[/math] (6) Does that seems correct? * from Minkowski formula here we have [math]x=D[/math] [math]y=0[/math] [math]z=0[/math] [math]t=T[/math] which give [math]S^2=D^2 - C^2T^2[/math] introducing (1) [math]S^2=C^2 T^2 - C^2T^2[/math] which has a null result. Edited December 19, 2010 by michel123456 Link to comment Share on other sites More sharing options...

Ludwik Posted December 21, 2010 Share Posted December 21, 2010 I am not convinced that your equation (4) is correct. Link to comment Share on other sites More sharing options...

swansont Posted December 21, 2010 Share Posted December 21, 2010 Equation 2 and 3 are wrong; you have a problem with your units — you are mixing distance units and time units S=0 implies a lightlike separation, i.e. the points are ct apart. Link to comment Share on other sites More sharing options...

michel123456 Posted December 21, 2010 Author Share Posted December 21, 2010 (edited) Equation 2 and 3 are wrong; you have a problem with your units you are mixing distance units and time units Yes, I am mixing distance & time. that doesn't make the equation wrong, that makes it unsolvable. S=0 implies a lightlike separation, i.e. the points are ct apart. Yes. That is the case in this rod experiment. The rod do not move. X on the diagram is a lightlike displacement. I am not convinced that your equation (4) is correct. Neither am I. that is the reason I put a "if" before the equation. And it is also the main reason of my question "Does that seems correct?" The bizarre here is that we have in the diagram a lightlike interval which is represented as an obvious quantity X, on one part, and on the other part we have to follow the statement that it must be null. The only way I can imagine to solve ambiguity is to say that the 2 point of vues are as viewed from different Frames Of Reference. The diagram (fig.1) is from a timeless FOR, contrarily to the standard equation (S^2=0) which is from a timed FOR. In this last case, it means that we, as observators, have moved in time the interval T. Which IMHO makes some sense. Edited December 21, 2010 by michel123456 -1 Link to comment Share on other sites More sharing options...

swansont Posted December 21, 2010 Share Posted December 21, 2010 Yes, I am mixing distance & time. that doesn't make the equation wrong, that makes it unsolvable. It can describe no physical situation. "Unsolvable" or "wrong" is just semantics at that point. Nothing based on those equations will have any valid meaning. Link to comment Share on other sites More sharing options...

michel123456 Posted December 21, 2010 Author Share Posted December 21, 2010 (edited) It can describe no physical situation. "Unsolvable" or "wrong" is just semantics at that point. Nothing based on those equations will have any valid meaning. You must agree that the diagram represents something. The equation (2) is simply a pythagorean triangle. There is nothing wrong in it. Edited December 21, 2010 by michel123456 -1 Link to comment Share on other sites More sharing options...

swansont Posted December 21, 2010 Share Posted December 21, 2010 Yes it is a triangle. But you are trying to do some physics with it, and in that context, we insist that the units always be the same in an equation. You have drawn what would be part of the light cone for this system. http://en.wikipedia.org/wiki/Light_cone Link to comment Share on other sites More sharing options...

michel123456 Posted December 22, 2010 Author Share Posted December 22, 2010 Yes it is a triangle. But you are trying to do some physics with it, and in that context, we insist that the units always be the same in an equation. You have drawn what would be part of the light cone for this system. http://en.wikipedia.org/wiki/Light_cone Exactly. equation (4) stands only for a time slice in present so that δΤ=0 In this case, for both timeless and timed observators equation (6) must be correct. Link to comment Share on other sites More sharing options...

swansont Posted December 22, 2010 Share Posted December 22, 2010 But [math]C^2\neq-1[/math] The speed of light is not an imaginary number. Link to comment Share on other sites More sharing options...

michel123456 Posted December 22, 2010 Author Share Posted December 22, 2010 (edited) But [math]C^2\neq-1[/math] The speed of light is not an imaginary number. here the speed of light is an angle. The expected result was C=1. It does not come out. Edited December 22, 2010 by michel123456 Link to comment Share on other sites More sharing options...

swansont Posted December 22, 2010 Share Posted December 22, 2010 That's because your setup has no physical meaning. c isn't an angle, x isn't a distance Link to comment Share on other sites More sharing options...

michel123456 Posted December 22, 2010 Author Share Posted December 22, 2010 That's because your setup has no physical meaning. c isn't an angle, x isn't a distance In all spacetime diagrams speed is an angle. Link to comment Share on other sites More sharing options...

swansont Posted December 22, 2010 Share Posted December 22, 2010 Speed is represented by an angle. Link to comment Share on other sites More sharing options...

DrRocket Posted February 20, 2011 Share Posted February 20, 2011 (edited) Let's take a rod. A hollow cylinder. Let's put it horizontaly, with a laser device at one end. The rod is standing still at rest. Let's open the laser for a time T so that light travels inside the cylinder a distance D. In Spacetime, light goes from point A to B, having travelled the interval X. Something like fig.1 here below: In this diagram, speed of light, C, is the angle of the line AB. It is expressed as a fraction: meters/seconds. Because we are measuring light, we can pose that [math] D=CT [/math] (1) From the diagram, using Pythagoream theorem, we can pose that [math] X^2 = D^2 + T^2 [/math] (2) Replacing (1) in (2) we get [math] X^2 = C^2 T^2 + T^2 [/math] (3) From the diagram, it appears that X corresponds to the definition of the Minkowski spacetime interval S. Because it is a lightlike spacetime interval, S=0* If we pose that X (from the diagram) is the same as S (from Theory) we get [math] X^2 = C^2 T^2 + T^2=S=0[/math] (4) or [math]C^2 T^2 + T^2=0[/math] (5) which has the only solution [math]C^2=-1[/math] (6) Does that seems correct? * from Minkowski formula here we have [math]x=D[/math] [math]y=0[/math] [math]z=0[/math] [math]t=T[/math] which give [math]S^2=D^2 - C^2T^2[/math] introducing (1) [math]S^2=C^2 T^2 - C^2T^2[/math] which has a null result. The objection to "mixing units" is a red herring. The constancy of c allows distance and time to be expressed in the same units. The "mixing" of time and space is the whole point of spacetime. Forget specific coordinates for the moment. Any timelike unit vector can be the time axis for some observer. You have to be able to handle time and space on the same footing. The problem here is that you are using the Euclidean metric (equations 2, 3) in a Minkowskian space. Equation 4 which basically asserts the equality of the Euclidean and Minkowski metrics (you left out an exponent but that is not important) is just flat wrong -- this is the critical mistake. You need to be consistent and use the Minkowski metric (what you called the Minkowski formula) everywhere. It is the metric that determines the geometry. You can successfully mix neither metrics nor metaphors. See sections 1.4 and 1.5 in Gravitation by Misner, Thorne, and Wheeler -- here. Edited February 20, 2011 by DrRocket Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Author Share Posted February 20, 2011 I think my diagram ressembles a spacetime diagram: space on abscissa & time on ordinate. How could I state differently equations (2) and (3) in Minkowskian space? Link to comment Share on other sites More sharing options...

DrRocket Posted February 20, 2011 Share Posted February 20, 2011 I think my diagram ressembles a spacetime diagram: space on abscissa & time on ordinate. How could I state differently equations (2) and (3) in Minkowskian space? You can't. That is pretty much the point of the Minkowski metric. It is also the wrong question. When you are doing abstract mathematics you need to ask the question "Why can I do this ?" before you do it, not "Why can't I do that?" afterward. Your equations 2 and 3 purport to establish a relationship between position/distance and time in what is really a Newtonian perspective. But that relationship is basically speed, and speed is not universal, except for light. So no such relationship should be expected to exist. The difference with the Minkowski metric lies in the relationship between Minkowski distance and time -- Minkowski distance is proper time (see the thread on proper time). The result is that all 4-velocities, not just for light, are c (usually in relativity one chooses units in which c=1). That is why one can relate distance and time via the Minkowski metric, while you cannot do that via the Euclidean metric. You are trying to understand special relativity and the geometry of Minkowski space, fairly abstract concepts, with a Newtonian and Euclidean mindset. Applying "common sense" will not work. That is a path that is fraught with peril. I suggest finding a copy of Gravitation by Misner, Thorne, and Wheeler, or using the link in my earlier post, and reading Chapter 1 very carefully, several times, to get the coordinate-free perspective of spacetime. Link to comment Share on other sites More sharing options...

michel123456 Posted February 20, 2011 Author Share Posted February 20, 2011 (edited) You can't. That is pretty much the point of the Minkowski metric. It is also the wrong question. When you are doing abstract mathematics you need to ask the question "Why can I do this ?" before you do it, not "Why can't I do that?" afterward. Your equations 2 and 3 purport to establish a relationship between position/distance and time in what is really a Newtonian perspective. But that relationship is basically speed, and speed is not universal, except for light. So no such relationship should be expected to exist. (...) Emphasis mine. The diagram is about light. And the angle represents speed, as you stated, which is C. (in orange color on the diagram) Edited February 20, 2011 by michel123456 Link to comment Share on other sites More sharing options...

DrRocket Posted February 20, 2011 Share Posted February 20, 2011 (edited) Let's take a rod. A hollow cylinder. Let's put it horizontaly, with a laser device at one end. The rod is standing still at rest. Let's open the laser for a time T so that light travels inside the cylinder a distance D. In Spacetime, light goes from point A to B, having travelled the interval X. Something like fig.1 here below: In this diagram, speed of light, C, is the angle of the line AB. It is expressed as a fraction: meters/seconds. Because we are measuring light, we can pose that [math] D=CT [/math] (1) From the diagram, using Pythagoream theorem, we can pose that [math] X^2 = D^2 + T^2 [/math] (2) Replacing (1) in (2) we get [math] X^2 = C^2 T^2 + T^2 [/math] (3) From the diagram, it appears that X corresponds to the definition of the Minkowski spacetime interval S. Because it is a lightlike spacetime interval, S=0* If we pose that X (from the diagram) is the same as S (from Theory) we get [math] X^2 = C^2 T^2 + T^2=S=0[/math] (4) or [math]C^2 T^2 + T^2=0[/math] (5) which has the only solution [math]C^2=-1[/math] (6) Does that seems correct? * from Minkowski formula here we have [math]x=D[/math] [math]y=0[/math] [math]z=0[/math] [math]t=T[/math] which give [math]S^2=D^2 - C^2T^2[/math] introducing (1) [math]S^2=C^2 T^2 - C^2T^2[/math] which has a null result. Emphasis mine. The diagram is about light. And the angle represents speed, as you stated, which is C. (in orange color on the diagram) OK, let's walk through the original logic: "Because we are measuring light, we can pose that [math]D=CT [/math] (1)" True, but simply because distance = rate x time "From the diagram, using Pythagoream theorem, we can pose that [math] X^2 = D^2 + T^2 [/math] (2)" This is really just a definition of [math]X[/math] as a point in an artifical Euclidean space of Time x Space, but spacetime is not just Time X Space. A map of Delaware and a map of Beirut dan be laid out with the same numerical grid system, but Delaware is not Beirut. "Replacing (1) in (2) we get [math] X^2 = C^2 T^2 + T^2 [/math] (3)" True, but artificial as noted above. "If we pose that X (from the diagram) is the same as S (from Theory) we get [math] X^2 = C^2 T^2 + T^2=S=0[/math] (4)" [math]X[/math] has absolutely nothing to do with [math]S[/math] so you cannot "pose" this. This is the source of the mistake. But it is a profound mistake. Everything above might be technically true, but it is also trivial. All of the conclusions that are subsequently drawn are based on this false premise. Edited February 22, 2011 by DrRocket Link to comment Share on other sites More sharing options...

michel123456 Posted February 21, 2011 Author Share Posted February 21, 2011 _Pose is suppose without sup. _I am in a deep trouble since it is a profound mistake. _I agree there is a conceptional issue when equalizing [math] X [/math] with [math] S [/math]: [math] S [/math] is zero, and in the diagram [math] X [/math] is obviously not null. _Trivial is the answer I first thought would appear, that is C=1. Link to comment Share on other sites More sharing options...

md65536 Posted February 22, 2011 Share Posted February 22, 2011 (edited) In all spacetime diagrams speed is an angle. How do you get that speed is an angle? According to your diagram with c as an angle you get tan( c ) = T/D (opposite over adjacent). I think you want that c = D/T... a speed... the distance D that light travels in time T, such that c = tan( pi/2 - theta ) where theta is the angle you currently label as c. Or simply c = tan( theta ) if you use the angle off of the time axis rather than the distance axis. The tangent of an angle is not an angle. In this diagram, speed is a ratio. I don't know enough about spacetime diagrams, but if they do represent speed with a proper angle, I don't think this diagram corresponds to them (as you do to get eq 4). In Minkowski diagrams, the vertical axis is "ct", which is the same units as the distance axis, so uh... well I dunno, but I guess you can do different things with it and have it make sense, where the same thing wouldn't make the same sense with your diagram. Edited February 22, 2011 by md65536 Link to comment Share on other sites More sharing options...

michel123456 Posted February 22, 2011 Author Share Posted February 22, 2011 (edited) How do you get that speed is an angle? According to your diagram with c as an angle you get tan( c ) = T/D (opposite over adjacent). I already standed corrected by Swansont who answered immediately: Speed is represented by an angle. I never spoke about tan, one could use cotangent instead, but you are right. In my original sketch, rods were vertical & time was going horizontaly from left to right (like a comic strip). When I putted time vertical to make it look like a conventional space-time diagram, I forgot to switch the indication of the angle. You may notice that the angle is not 45 degrees, that is intentionally because as you wrote the time axis is not ct as in a Minkowski diagram. But it is still a space-time diagram. A very simple one, elementary one could say. I don't understand the reason why to call it an "artificial" one. Edited February 22, 2011 by michel123456 Link to comment Share on other sites More sharing options...

michel123456 Posted October 5, 2016 Author Share Posted October 5, 2016 (edited) This is the revival of an old thread.Looking at this below, from a post of Celeritas in another thread I wonder what is the fundamental difference between my original diagram and the one above? This one is about a rod that has been moved in space, and Euclidian geometry at work. But time is definitely there. My diagram is about a rod at rest, "moved" in time, and I was told that Euclidian geometry cannot work. But it looks to me that both diagrams are very similar, simply changed by a rotation. Why? Here a re-post of my original diagram from post #1 Edited October 5, 2016 by michel123456 Link to comment Share on other sites More sharing options...

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